如何在Node.js中获取脚本的路径?
我知道有流程。Cwd,但它只引用调用脚本的目录,而不是脚本本身。例如,假设我在/home/kyle/目录下,然后运行以下命令:
node /home/kyle/some/dir/file.js
如果我调用process.cwd(),我会得到/home/kyle/,而不是/home/kyle/some/dir/。有办法得到那个目录吗?
如何在Node.js中获取脚本的路径?
我知道有流程。Cwd,但它只引用调用脚本的目录,而不是脚本本身。例如,假设我在/home/kyle/目录下,然后运行以下命令:
node /home/kyle/some/dir/file.js
如果我调用process.cwd(),我会得到/home/kyle/,而不是/home/kyle/some/dir/。有办法得到那个目录吗?
当前回答
如果你使用pkg来打包你的应用,你会发现这个表达式很有用:
appDirectory = require('path').dirname(process.pkg ? process.execPath : (require.main ? require.main.filename : process.argv[0]));
process.pkg tells if the app has been packaged by pkg. process.execPath holds the full path of the executable, which is /usr/bin/node or similar for direct invocations of scripts (node test.js), or the packaged app. require.main.filename holds the full path of the main script, but it's empty when Node runs in interactive mode. __dirname holds the full path of the current script, so I'm not using it (although it may be what OP asks; then better use appDirectory = process.pkg ? require('path').dirname(process.execPath) : (__dirname || require('path').dirname(process.argv[0])); noting that in interactive mode __dirname is empty. For interactive mode, use either process.argv[0] to get the path to the Node executable or process.cwd() to get the current directory.
其他回答
我又看了一遍文件后找到的。我要找的是__filename和__dirname模块级变量。
__filename是当前模块的文件名。这是当前模块文件的解析绝对路径。(例:/ home /凯尔/一些/ dir / file.js) __dirname是当前模块的目录名。(例:/ home /凯尔/一些/ dir)
这个命令返回当前目录:
var currentPath = process.cwd();
例如,使用路径读取文件:
var fs = require('fs');
fs.readFile(process.cwd() + "\\text.txt", function(err, data)
{
if(err)
console.log(err)
else
console.log(data.toString());
});
Index.js中包含要导出的模块的任何文件夹
const entries = {};
for (const aFile of require('fs').readdirSync(__dirname, { withFileTypes: true }).filter(ent => ent.isFile() && ent.name !== 'index.js')) {
const [ name, suffix ] = aFile.name.split('.');
entries[name] = require(`./${aFile.name}`);
}
module.exports = entries;
这将找到当前目录的根目录下的所有文件,要求并导出与文件名干相同的导出名称的每个文件。
基本上你可以这样做:
fs.readFile(path.resolve(__dirname, 'settings.json'), 'UTF-8', callback);
使用resolve()而不是连接'/'或'\',否则您将遇到跨平台问题。
注意:__dirname是模块或包含脚本的本地路径。如果你正在编写一个插件,需要知道主脚本的路径,它是:
require.main.filename
或者,获取文件夹名称:
require('path').dirname(require.main.filename)
如果你使用pkg来打包你的应用,你会发现这个表达式很有用:
appDirectory = require('path').dirname(process.pkg ? process.execPath : (require.main ? require.main.filename : process.argv[0]));
process.pkg tells if the app has been packaged by pkg. process.execPath holds the full path of the executable, which is /usr/bin/node or similar for direct invocations of scripts (node test.js), or the packaged app. require.main.filename holds the full path of the main script, but it's empty when Node runs in interactive mode. __dirname holds the full path of the current script, so I'm not using it (although it may be what OP asks; then better use appDirectory = process.pkg ? require('path').dirname(process.execPath) : (__dirname || require('path').dirname(process.argv[0])); noting that in interactive mode __dirname is empty. For interactive mode, use either process.argv[0] to get the path to the Node executable or process.cwd() to get the current directory.